Rational Exponents: Practice

Handle the minus sign by taking the reciprocal, then work on the fraction.

$81^{-3/4} = \frac{1}{81^{3/4}} = \frac{1}{\left(\sqrt[4]{81}\right)^{3}} = \frac{1}{3^{3}} = \frac{1}{27}$

The base is positive, so the answer is positive. A negative exponent inverts the value rather than negating it.

Raising a power to a power multiplies the exponents.

$\left(x^{2/3}\right)^{9} = x^{\frac23 \cdot 9} = x^{18/3} = x^{6}$

The base is positive, which is exactly the hypothesis under which the power law was proved for rational exponents.

Dividing powers of the same base subtracts the exponents.

$\frac{y^{3/4}}{y^{1/4}} = y^{\frac34 - \frac14} = y^{2/4} = y^{1/2} = \sqrt{y}$

The tempting wrong move is to divide the exponents ($\tfrac34 \div \tfrac14 = 3$), but division of powers subtracts them.

First find the value: $16^{3/4} = \left(\sqrt[4]{16}\right)^{3} = 2^{3} = 8$. Now convert each expression. Both $\left(\sqrt[4]{16}\right)^{3}$ and $\sqrt[4]{16^{3}}$ equal $16^{3/4}$, since the two routes agree for a positive base, and $2^{3}$ is that same $8$.

$\sqrt[3]{16^{4}} = 16^{4/3} \neq 16^{3/4}$

Only $\sqrt[3]{16^{4}}$ swaps the root and the power, giving the exponent $\tfrac43$ instead of $\tfrac34$, so it is the one that is not equal.

Write the radical as a rational exponent, then multiply the exponents.

$\left(\sqrt[3]{5}\right)^{6} = \left(5^{1/3}\right)^{6} = 5^{\frac13 \cdot 6} = 5^{2} = 25$

Six copies of the cube root pair up into two copies of $5$, which is the same statement.

Multiplying powers of one base adds the exponents. Put the fractions over the common denominator $12$.

$a^{2/3} \cdot a^{3/4} = a^{\frac{8}{12} + \frac{9}{12}} = a^{17/12}$

Multiplying the exponents would give $a^{1/2}$, and adding numerators and denominators separately would give the meaningless $a^{5/7}$. Neither is the product law.

A negative exponent on a fraction flips the fraction over, which turns the exponent positive.

$\left(\frac{1}{16}\right)^{-3/4} = 16^{3/4} = \left(\sqrt[4]{16}\right)^{3} = 2^{3} = 8$

The base is less than $1$, so a negative exponent makes the answer larger than $1$.

Convert both radicals to exponents, then add, since the base is the same.

$\sqrt{x} \cdot \sqrt[3]{x} = x^{1/2} \cdot x^{1/3} = x^{\frac36 + \frac26} = x^{5/6}$

The conversion to exponents is what makes this easy. In radical form the two indices $2$ and $3$ have no obvious way to combine.

Inside the bracket, dividing subtracts the exponents, and subtracting a negative adds.

$\frac{x^{1/2}}{x^{-1/4}} = x^{\frac12 - \left(-\frac14\right)} = x^{\frac12 + \frac14} = x^{3/4}$

Now raise that to the fourth power, which multiplies the exponent by $4$.

$\left(x^{3/4}\right)^{4} = x^{3}$

The common slip is $\tfrac12 - \tfrac14 = \tfrac14$, which would give $x^{1}$. The sign of the exponent in the denominator matters.

Take the reciprocal for the minus sign, then the cube root, then the square.

$125^{-2/3} = \frac{1}{125^{2/3}} = \frac{1}{\left(\sqrt[3]{125}\right)^{2}} = \frac{1}{5^{2}} = \frac{1}{25}$

A positive base can never produce a negative power, no matter what the exponent is.

The denominator $2$ calls for a square root, so root first and then raise to the fifth.

$9^{5/2} = \left(\sqrt{9}\right)^{5} = 3^{5} = 243$

Forgetting the root entirely gives $9^{5} = 59049$, and treating the exponent as a multiplier gives $9 \cdot 2.5 = 22.5$. Both are traps.

Distribute the exponent over the product, multiplying each inner exponent by $-\tfrac12$.

$\left(x^{4}y^{-6}\right)^{-1/2} = x^{4 \cdot \left(-\frac12\right)}\,y^{-6 \cdot \left(-\frac12\right)} = x^{-2}y^{3}$

Now move the negative exponent to the denominator.

$x^{-2}y^{3} = \frac{y^{3}}{x^{2}}$

The two exponents end up with opposite signs, so one variable rises and the other falls.